Download Document

MATH 103A: MODERN ALGEBRA I (WINTER 2014)
HW8 SOLUTIONS
Problem 1. Let φ be a surjective homomorphism from S4 onto Z2 . By the First Homomorphism
Theorem, S4 /Ker (φ) ≈ φ(S4 ) = Z2 . So |S4 /Ker (φ)| = |Z2 | = 2. This means that Ker (φ) is a
normal subgroup of S4 with order 12. Hence, Ker (φ) = A4 .
The two homomorphisms from S4 to Z2 are:
• φ0 (σ) = 0, for all σ ∈ S4
(
0 for σ ∈ A4
• φ1 (σ) =
1 otherwise
Problem 2. Let G and H be groups and let α, β : G → H be homomorphism. Let K = { g ∈ G :
α( g) = β( g)}. Then K is a subgroup of G, following from the one-step subgroup test below.
• Since α(e) = e = β(e), e ∈ K. So, K 6= ∅.
• Let g, h ∈ K then
α( gh−1 ) = α( g)α(h−1 ) = α( g)α(h)−1 = β( g) β(h)−1 = β( g) β(h−1 ) = β( gh−1 ).
So, gh−1 ∈ K.
Problem 3. Suppose H and K are normal subgroups of G and H ∩ K = {e}. Consider the map
φ : G → G/H ⊕ G/K defined by φ( g) = ( gH, gK ). Then for any x, y ∈ G,
φ( xy) = ( xyH, xyK ) = ( xH, xK )(yH, yK ) = φ( x )φ(y).
Thus, φ is a homomorphism with Ker (φ) = { g ∈ G |( gH, gK ) = ( H, K )} = H ∩ K = {e}.
By the First Isomorphism Theorem, G = G/{e} = G/Ker (φ) ≈ φ( G/H ⊕ G/K ) ⊆ G/H ⊕ G/K.
Problem 4. Let p be a prime number. For each pair ( x, y) ∈ Z2p , let φ( x,y) : Z p ⊕ Z p → Z p be a
mapping such that φ( x,y) (1, 0) = x and φ( x,y) (0, 1) = y, for x, y ∈ Z p . For any ( a, b) ∈ Z p ⊕ Z p ,
define φ( x,y) ( a, b) = ax + by.
If ( a, b) = (c, d) then p|( a − c) and p|(b − d). Thus, p|( ax + by − cx − dy). Hence, φ( x,y) is welldefined. Furthermore,
φ( x,y) (( a, b) + (c, d)) = ( a + c) x + (b + d)y = ( ax + by) + (cx + dy) = φ( x,y) ( a, b) + φ( x,y) (c, d).
Thus, φ( x,y) is a homomorphism, for each ( x, y) ∈ Z2p . Therefore, the number of homomorphisms
from Z p ⊕ Z p to Z p is p2 .
1
Problem 5. Suppose φ : G → Z6 ⊕ Z2 is a surjective homomorphism and |Ker (φ)| = 5. Since
Z6 ⊕ Z2 is Abelian, every subgroup of the group is normal. So, Z6 ⊕ Z2 has normal subgroups
of orders 1, 2, 3, 4, 6, and 12.
By Theorem 10.2.8, if K / Z6 ⊕ Z2 , then φ−1 (K ) / G. Now by Theorem 10.2.5, since |Ker (φ)| =
5, |φ−1 (K )| = 5|K |. Therefore, G has normal subgroups of orders 5, 10, 15, 20, 30, and 60.
Problem 6. In Z16 , there are:
• 1 element of order 2: 8
• 2 elements of order 4: 4 and 12
In Z8 ⊕ Z2 , there are:
• 3 element of order 2: (4, 0), (4, 1), and (0, 1)
• 4 elements of order 4: (2, 0), (6, 0), (2, 1), , and (6, 1)
In Z4 ⊕ Z2 ⊕ Z2 , there are:
• 7 element of order 2: (2, 0, 0), (2, 0, 1), (2, 1, 0), (2, 1, 1), (0, 0, 1), (0, 1, 0) and (0, 1, 1)
• 8 elements of order 4: (1, 0, 0), (3, 0, 0), (1, 0, 1), (3, 0, 1), (1, 1, 0), (3, 1, 0), (1, 1, 1), and (3, 1, 1)
Problem 7. Let p1 , p2 , . . . , pn be distinct primes and let G be an Abelian group of order p41 p42 . . . p4n .
Then G ≈ G1 ⊕ G2 ⊕ · · · ⊕ Gn , where Gi is an Abelian group of order p4i , for all i ∈ {1, 2, . . . , n}.
For each i ∈ {1, 2, , . . . , n}, there are five distinct direct products for Gi , as demonstrated in the
textbook. Therefore, in total, there are 5n Abelian groups of order p41 p42 . . . p4n .
Problem 8. The set G = {1, 4, 11, 16, 19, 26, 29, 31, 34, 41, 44} is a group under multiplication mod
45. Since G is Abelian with | G | = 12, we know that G is isomorphic to either Z12 or Z6 ⊕ Z2 .
Further calculation shows that G does not have any element of order 12. Thus, G ≈ Z6 ⊕ Z2 .
Observe that 12 = 22 3 where 2 is a prime that does not divide 3. By Lemma 1 (page 231), G =
2
H × K where H = { x ∈ G | x2 = e} = {1, 19, 26, 44} = h19i × h26i, and K = { x ∈ G | x3 = e} =
{1, 16, 31} = h16i. So G = h19i × h26i × h16i.
Problem 9. Let G be the group of all n × n diagonal matrix with ±1 diagonal entries. Then
| G | = 2n and every non-identity element of G has order 2. Thus, G ≈ Z2 × Z2 × · · · × Z2 .
2